Question

1.00 kg of ice at -10 °C is heated using a Bunsen burner flame until all the ice melts and the temperature reaches 95 °C. A) How much energy in kJ is required to effect this transformation?

Answer 1

Answer : The energy required is, 574.2055 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(95^oC)$

Now we have to calculate the enthalpy change or energy.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = energy required = ?

m = mass of ice = 1 kg  = 1000 g

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

n = number of moles of ice = $\frac{\text{Mass of ice}}{\text{Molar mass of ice}}=\frac{1000g}{18g/mole}=55.55mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC]$

$\Delta H=574205.5J=574.2055kJ$     (1 KJ = 1000 J)

Therefore, the energy required is, 574.2055 KJ